2- This reaction is endothermic, as it is absorbing the heat it needs from the surrounding water. This cools the water down.
2nd Link
1- a) C + O2 -> CO2 ∆ H = -393.5 kJ
b) Cu + Cl2 -> CuCl2 ∆ H = -220.1 kJ
c) Cu + ½ Cl2 -> CuCl ∆ H = -137.2 kJ
d) N2 + 2H2 -> N2H4 ∆ H = 50.6 kJ
e) N2 + 2H2 + ½ Cl2 -> N2H4Cl ∆ H = -314.4 kJ
2- Molar Mass of SO2 is molar mass of S + 2O.
Molar Mass is 64.1 g/mol.
25.0 g * 1 mol/64.1 g = 0.39 mol
Hf = -296.8 kJ/mol
E = -296.8 * 0.39 mol
E = -115.8 kJ
3- 11.5 g * 1 mol/46.1 g = 0.25 mol
Hr = -950 kJ/mol
E = 950 kJ/mol* 0.25 mol
E = 237.5 kJ
4- 33.0 g * 1 mol/44.1 g = 0.75 mol
Hr = -2220 kJ/mol
E = -2220kJ/mol* 0.75 mol
E = -1661 kJ
Assignment:
1.) STP – Standard Temperature and Pressure - 0oC (273.15 K, 101.325 kPa)
It’s commonly used to define standard conditions for temperature and pressure which is important for the measurements of chemical or physical processes.
SATP – Standard Ambient Temperature and Pressure - 25oC (298.15 K,101 kPa)
It is used as a reference in chemistry.
When scientists are experimenting with gases, they try to work in the same conditions (temperature and pressure) or their results would be different. They, therefore, develop standard conditions when working with gases.
2.) a. 2 Hg(l) + I2 (s) ® Hg2I2(s) DH° = -28.9 kcal
Exothermic: 2 Hg(l) + I2 (s) ® Hg2I2(s) + 28.9 kcal
d. Na(s) ® Na (g) DH° = 25.98 kcal
Endothermic: Na(s) + 25.98 kcal ® Na (g)
3.) Graph 1 = Exothermic as the potential energy in its system is now released into the environment
Graph 2 = Endothermic as it has absorbed the energy into its system.
b.) Graph 1 would have a negative value for DH as it is exothermic.
4.) a. SO2(g)
S + 02 à SO2 + 296.8kJ
b. C3H8 (g)
3C + 4H2 à C3H8 + 103.8 kJ
c. N2O (g)
N2 + 1/2O2 + 82.1kJ à N2O (g)
d. Na2CO3 (s)
2Na + C + 3/2O3 à Na2CO3 (s) + 1130.7kJ
5.) A D and F would be spontaneous as they are all exothermic.
1- a) Exothermic
b) Endothermic
c) Exothermic
2- This reaction is endothermic, as it is absorbing the heat it needs from the surrounding water. This cools the water down.
2nd Link
1- a) C + O2 -> CO2 ∆ H = -393.5 kJ
b) Cu + Cl2 -> CuCl2 ∆ H = -220.1 kJ
c) Cu + ½ Cl2 -> CuCl ∆ H = -137.2 kJ
d) N2 + 2H2 -> N2H4 ∆ H = 50.6 kJ
e) N2 + 2H2 + ½ Cl2 -> N2H4Cl ∆ H = -314.4 kJ
2- Molar Mass of SO2 is molar mass of S + 2O.
Molar Mass is 64.1 g/mol.
25.0 g * 1 mol/64.1 g = 0.39 mol
Hf = -296.8 kJ/mol
E = -296.8 * 0.39 mol
E = -115.8 kJ
3- 11.5 g * 1 mol/46.1 g = 0.25 mol
Hr = -950 kJ/mol
E = 950 kJ/mol* 0.25 mol
E = 237.5 kJ
4- 33.0 g * 1 mol/44.1 g = 0.75 mol
Hr = -2220 kJ/mol
E = -2220kJ/mol* 0.75 mol
E = -1661 kJ
Assignment:
1.) STP – Standard Temperature and Pressure - 0oC (273.15 K, 101.325 kPa)
It’s commonly used to define standard conditions for temperature and pressure which is important for the measurements of chemical or physical processes.
SATP – Standard Ambient Temperature and Pressure - 25oC (298.15 K,101 kPa)
It is used as a reference in chemistry.
When scientists are experimenting with gases, they try to work in the same conditions (temperature and pressure) or their results would be different. They, therefore, develop standard conditions when working with gases.
2.) a. 2 Hg(l) + I2 (s) ® Hg2I2(s) DH° = -28.9 kcal
Exothermic: 2 Hg(l) + I2 (s) ® Hg2I2(s) + 28.9 kcal
b. N2 (g) + 3 F2 (g) ® 2 NF3 (g) + 27.2 kcal
Exothermic: N2 (g) + 3 F2 (g) ® 2 NF3 (g) DH° = -27.2 kcal
c. NH4NO3(s) + 6.1 kcal ® NH4+ (aq) + NO3- (aq)
Endothermic: NH4NO3(s) ® NH4+ (aq) + NO3- (aq) DH° = +6.1 kcal
d. Na(s) ® Na (g) DH° = 25.98 kcal
Endothermic: Na(s) + 25.98 kcal ® Na (g)
3.) Graph 1 = Exothermic as the potential energy in its system is now released into the environment
Graph 2 = Endothermic as it has absorbed the energy into its system.
b.) Graph 1 would have a negative value for DH as it is exothermic.
4.) a. SO2(g)
S + 02 à SO2 + 296.8kJ
b. C3H8 (g)
3C + 4H2 à C3H8 + 103.8 kJ
c. N2O (g)
N2 + 1/2O2 + 82.1kJ à N2O (g)
d. Na2CO3 (s)
2Na + C + 3/2O3 à Na2CO3 (s) + 1130.7kJ
5.) A D and F would be spontaneous as they are all exothermic.
Hess' Law:
1.) #1 - 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l) ΔH° = -1170 kJ
#2 - 4 NH3 (g) + 3 O2 (g) → 2 N2 (g) + 6 H2O (l) ΔH° = -1530 kJ
#3. - 1/2N2 + 1/202 --> NO
Reverse Equation #2
4NH3 + 5O2 --> 4NO + 6H2O ∆ H = -1170kJ
2N2 + 6H2O --> 4NH3 + 3O2 ∆ H = +1530kJ
Add the two equations and get the result:
2N2 + 2O2 --> 4NO
divide the coefficients by 4
1/2N2 +1/2 O2 --> NO DH = +90kJ